Htykuut

If p is the probability that a man aged x will die in a year then the
probability that out of n men $A_1, A_2 ,A_3 ...A_n$, each aged x,
$A_1$ will die in an year and be the first to die is: ?

a) $1- (1- p)^{n-1}$

b)$(1-p)^n$

c) $dfrac 1n [1- (1-p)^n]$

d) $dfrac 1n (1-p)^n$

Attempt:

Order of dying matters.

Choose $r$ out of $n-1$ and arrange them (i.e. their deaths) in $r!$ ways.

So $P(E) = ptimes{^{n-1}C_0} (1- p)^{n-1}+ p times ptimes {^{n-1}C_1} (1-p)^{n-1} + ptimes p^2 times 2!times {^{n-1}C_2} (1- p)^{n-2}+...$

$implies P(E) = p sum_{r=0}^{n-1} {^{n-1}}C_r (1-p)^{n-1-r}p^r times r!$

But this doesn't match any of the options.

Could someone please tell me where I have gone wrong?

Edit: My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate

Your expression is almost correct. Let $N$ be the number of deaths out of $A_2, ldots, A_n$. We know that $A_1$ must die. After that, we decompose the event with respect to $N$ (notice we are assuming all deaths are independent) and use that given $N = r$, the probability that the first to die is $A_1$ is $frac{1}{r+1}$ (here assuming that any order of deaths is equally likely). Thus:
begin{align*}
mathbb{P}[text{given event}]
&= mathbb{P}[A_1 text{dies}] cdot sum_{r=0}^{n-1} mathbb{P}[N=r] cdot mathbb{P}[A_1 text{dies first} mid N=r]\
&= p cdot sum_{r=0}^{n-1} binom{n-1}{r} p^r (1-p)^{n-1-r} cdot frac{1}{r+1}
end{align*}
To simplify the expression above, we notice that
$$
binom{n-1}{r}frac{1}{r+1} = frac1n binom{n}{r+1}.
$$
From the above, we can write
begin{align*}
mathbb{P}[text{given event}]
&= frac{1}{n} cdot sum_{r=0}^{n-1} binom{n}{r+1} p^{r+1} (1-p)^{n-1-r} \
&= frac1n [(p + (1-p))^n - (1-p)^n] = frac1n [1 - (1-p)^n].
end{align*}

They probability for a person aged $x$ to not die is $1-p$. Hence, the probability that none of them is $(1-p)^n$. Therefore, the probability that at least one of them will die is $1-(1-p)^n$. SInce every one of them has the same probability to be that first person (as all are identical), it being $A_1$ has probability ${1 over n}[1-(1-p)^n]$.

Hence, option $(c)$ is correct.